nayanparikh wrote:
If ab ≠ 0, is |a-b|>|a+b|?
1) ab < 0
2) a > b
Can any one please provide explanation on how to solve these type of questions ?
@Bunnel ?
1) ab < 0
Means that there are two possibilities: either a>0 and b<0 or a<0 and b>0. Let's look at htose two seperately.
a>0, b<0. Let's denote b=-y (y - some positive integer). -y<0, y>0.
|(a - (-y))| > |a - y|
|a + y| > |a - y|
On the left side we have sum of two positive integers so we can pull off our modulus. On the right side we have difference o two positive integers which will always be lesser than their sum.
a<0, b>0. Same approach. a=-x (x some positive integer). -x<0, x>0
|-x - b| > |-x + b|
| - (x +b)| > |b - x|
x + b > |b - x|
Again we have sum of two positive integers on the left and their diffrence on th right. Sufficient.
2) a > b
We have three possible scenarious.
a) a>0, b>0
b) a<0, b<0
c) a>0, b<0
a) To eveade confusion it would be impler to use particular numbers here.
a = 30, b = 10
|30 - 10| ... |30 + 10| ----> 20 < 40
b) a = -10 b = -30
|-10 - (-30)| ... | -10 - 30|
| -10 + 30| ... | - (10 + 30)|
20 < 40
c) In c we have same case as in (1)
a=10, b=-30
|10 - (-30)| ... |10 - 30|
40 > 20 Insufficient
Answer A
I hope my approach wasn't too vague. May be our experts will give you more details.